Capacitors in Circuit: Voltage, Charge, and Switch Behavior Explained

[diredragon]

Homework Statement

Homework Equations

3. The Attempt at a Solution [/B]
First i found the voltage $U_{13}=\frac{E(R2+R3)}{R1+R2+R3}=50V$
Then i figured that the charge on 1 and 2 is equal since the voltage across is the same right?
$Q1=Q2=U_{13}\frac{C1C2}{C1+C2}=80mC$ and $Q1=Q2=U_{13}\frac{C3C4}{C3+C4}=125mC$
The the voltage AB is $U_{ab}=\frac{Q3}{C3} - \frac{Q1}{C1}=15V$ Is this right?
Now i have a problem figuring out what happens when the switch 2 closes. What changes?

 

[cnh1995]

First i found the voltage U13=E(R2+R3)R1+R2+R3=50VU13=E(R2+R3)R1+R2+R3=50VU_{13}=\frac{E(R2+R3)}{R1+R2+R3}=50V
Then i figured that the charge on 1 and 2 is equal since the voltage across is the same right?
Q1=Q2=U13C1C2C1+C2=80mCQ1=Q2=U13C1C2C1+C2=80mCQ1=Q2=U_{13}\frac{C1C2}{C1+C2}=80mC and Q1=Q2=U13C3C4C3+C4=125mCQ1=Q2=U13C3C4C3+C4=125mCQ1=Q2=U_{13}\frac{C3C4}{C3+C4}=125mC
The the voltage AB is Uab=Q3C3−Q1C1=15VUab=Q3C3−Q1C1=15VU_{ab}=\frac{Q3}{C3} - \frac{Q1}{C1}=15V Is this right?

Right.

After switch 2 is closed, what will be the steady state current through the capacitors?

 

[diredragon]

I've been thinking if this would be a good way to do it:
Since there is now a wire in between the first two capacitors we have separate voltages while in the other 2 everything remains the same.
$U_{c1}=\frac{Q1}{C1}=10V$ [the first case]
$U_{c1}=\frac{R2}{R1+R2+R3}E=40V$ [the second case of voltage on first capacitor]
$U_{c2}=\frac{R3}{R1+R2+R3}E=10V$ [the second case of voltage on second capacitor
So the new $U_{ab}=\frac{Q3}{C3}-U_{c1}=-15V$ [the first part remains the same while the voltage on C1 is different.
So the change is $ΔU_{ab}=-30V$ [Is this correct thinking?]
As for the $qp2$ i guessed that is the amount of charge that goes through.
That could be found using the voltage across the first capacitor and across the second:
So: $U_{c1}=\frac{Q1+q1}{C1}$ and $U_{c2}=\frac{Q2+q2}{C2}$ So the total 1 should be the sum right?
$q=q1+q2$

 

[cnh1995]

I've been thinking if this would be a good way to do it:
Since there is now a wire in between the first two capacitors we have separate voltages while in the other 2 everything remains the same.
Uc1=Q1C1=10VUc1=Q1C1=10VU_{c1}=\frac{Q1}{C1}=10V [the first case]
Uc1=R2R1+R2+R3E=40VUc1=R2R1+R2+R3E=40VU_{c1}=\frac{R2}{R1+R2+R3}E=40V [the second case of voltage on first capacitor]
Uc2=R3R1+R2+R3E=10VUc2=R3R1+R2+R3E=10VU_{c2}=\frac{R3}{R1+R2+R3}E=10V [the second case of voltage on second capacitor
So the new Uab=Q3C3−Uc1=−15VUab=Q3C3−Uc1=−15VU_{ab}=\frac{Q3}{C3}-U_{c1}=-15V [the first part remains the same while the voltage on C1 is different.
So the change is ΔUab=−30VΔUab=−30VΔU_{ab}=-30V [Is this correct thinking?]

Yes.

That could be found using the voltage across the first capacitor and across the second:
So: Uc1=Q1+q1C1Uc1=Q1+q1C1U_{c1}=\frac{Q1+q1}{C1} and Uc2=Q2+q2C2Uc2=Q2+q2C2U_{c2}=\frac{Q2+q2}{C2} So the total 1 should be the sum right?
q=q1+q2

Calculate the charge on each capacitor (using Q=CV) before and after the switch is closed. The difference between the two charges is the charge flowed through the switch.

 

[Delta2]

Though the problem doesn't ask us to bother with what exactly is happening in the transient state after switch 2 closes, I think it worth considering what happens there. I believe that the "naïve" picture that the capacitors C3 and C4 do not participate by charging or discharging during that stage does not hold. My main argument is that during the transient stage 2 the current through R1 is time varying (correct or not) hence the voltage on the edges of C3 and C4 also changes which means they are charging or discharging... Do you find this correct?

 

[cnh1995]

My main argument is that during the transient stage 2 the current through R1 is time varying (correct or not) hence the voltage on the edges of C3 and C4 also changes which means they are charging or discharging... Do you find this correct?

C3 and C4 are not charging or discharging.
They maintain the initial voltage of 25V each.

 

[epenguin]

"Then i figured that the charge on 1 and 2 is equal " that's right but "since the voltage across is the same right?“ is wrong.

The voltage across (across what?) is not the same. The reason the charge on 1 is the same as that on 2 is not "the voltage is the same" whatever that means, it is that the charges on two or any number of capacitors in simple series is always the same. Do some revision on that, or else I have discussed it in several posts in the last months.

They have given you simple numbers to 'help' you, but maybe also to trap you. Maybe you have fallen into the trap. The resistances are in a simple 4:1 ratio, And so are the capacitances, of the leftmost capacitors branch. So you're invited to think aha! the voltages at the two junction points are the same, there will be no current when I close switch 2. With switch open you have a 40V and a 10V voltage drop in both the resistor and capacitor branches - but they are at the opposite right way round aren't they?. Nevermind you can calculate the charges with switch open easily. Then after you close the switch and come to new equilibrium the voltage at the junction points will be the same as it is in the resistance branch when switch to it is Open. Open close this doesn't change so Nothing new to calculate about that. You can calculate now the new charges at equilibrium. You already have the voltages. As I think has already been said you don't need to calculate voltage at B, change of voltage between A and B will be the same as the change of voltage between A and anywhere, so you only need to use the change at A.
Reference https://www.physicsforums.com/threads/capacitors-in-a-circuit.899175/#post-5658733

 

[Delta2]

C3 and C4 are not charging or discharging.
They maintain the initial voltage of 25V each.

Ok fine can you or someone else do me a favour and present me with an analysis of what is exactly happening during the transient state after switch 2 is closed. I try to think of it but my brain goes into endless circles. The reason is that the initial (when switch 2 is closed) and final state (when the new equilibrium has reached) is the same with regards to the current that flows through R1 , R2 and R3. So how does the current during the transient, leaves that value and returns back to that value, isn't it a little mystery?

 

[cnh1995]

Ok fine can you or someone else do me a favour and present me with an analysis of what is exactly happening during the transient state after switch 2 is closed. I try to think of it but my brain goes into endless circles. The reason is that the initial (when switch 2 is closed) and final state (when the new equilibrium has reached) is the same with regards to the current that flows through R1 , R2 and R3. So how does the current during the transient, leaves that value and returns back to that value, isn't it a little mystery?

Current through R1 is 0.5A before and after switch 2 is closed. It doesn't change.
The voltages across C1 and C2 are changing instantaneously. This means if switch 2 has zero resistance, the currents through the capacitors are infinite in zero time.

When S2 is closed, two voltage sources of different voltage are connected in parallel. If they were batteries, it would be an invalid condition. But here, it works because one of them is a capacitor.
So, when S2 is closed, a momentary high current (ideally infinite) flows clockwise in the upper loop C1-R2-Π2 and another momentary high current flows through the lower loop C2-R3-Π2 anticlockwise. Both the currents add up in Π2.

Practically, there would be some very small resistance associated with the switch. So the currents will actually be current "pulses" with high magnitude and can cause sparking across the switch.

You can see the voltages across C3 and C4 remain unchanged after S2 is closed. Currents (pulses) only flow through the two middle loops.

 

[Delta2]

Current through R1 is 0.5A before and after switch 2 is closed. It doesn't change.
The voltages across C1 and C2 are changing instantaneously. This means if switch 2 has zero resistance, the currents through the capacitors are infinite in zero time.

When S2 is closed, two voltage sources of different voltage are connected in parallel. If they were batteries, it would be an invalid condition. But here, it works because one of them is a capacitor.
So, when S2 is closed, a momentary high current (ideally infinite) flows clockwise in the upper loop C1-R2-Π2 and another momentary high current flows through the lower loop C2-R3-Π2 anticlockwise. Both the currents add up in Π2.

Practically, there would be some very small resistance associated with the switch. So the currents will actually be current "pulses" with high magnitude and can cause sparking across the switch.

You can see the voltages across C3 and C4 remain unchanged after S2 is closed. Currents (pulses) only flow through the two middle loops.

Something doesn't look quite right with this analysis (infinite or very high currents that DO NOT pass through the real voltage source (current through R1 remains constant)...) but it might be possible. I ll take it Ok for now, I might post later. Thanks.

 

[epenguin]

You are quite right to try and get a proper physical account. Often lost sight of I think by students and perhaps in the teaching, in getting all the calculational detail complex circuits right.

This is just about to discharge of capacitors. Better revise what your books say about this subject. Best to start with simpler circuits.
For example of a battery, capacitor and resisteo in series. No current once the capacitor there is charged. If you instantaneously diminished the voltage ,the capacitor would partially discharge - some of its charge would flow through the resistor and battery. This is a case of no current flowing at the equilibrium state
Then imagine just the capacitor and resistor in parallel, a voltage across the two given by a battery and resistor in series, a bit more like your circuit but simplified. There then is a current flowing all the time. Then if you change instantaneously the applied voltage (or equivalently either of the resistors) charging the capacitor would redistribute itself. If you can get it clear for a simpler circuit I think you will have not much difficulty with the given one.

I agree with some of what cnh says, that's on closing the switch there is an instantaneous voltage jump,but not that the current on closing the switch is instantaneously infinite. The charge flows and redistributes itself at a rate that depends on the resistance in the circuit. It redistributes in time according to a negative exponential, that is the discharge rate gets slower and slower - as a matter of fact it is theoretically never 100% redistributed, though it is for all practical purposes.

 

[cnh1995]

I just ran a simulation and it is giving me a different result.
The transient after S2 is closed is different from what I thought in #9.
After S2 is closed, there is increase in I_R1, decrease in I_R2 and increase in I_R3. Therefore, the voltage across the capacitor branches decreases and then increases exponentially back to 50V.
Also, the current through the switch is not that high. I must have misunderstood something while drawing equivalent circuits of the given network.
Here's a screenshot showing the transient.

Green one is the current through R1, orange one is the switch current and blue one is the capacitor branch voltage.

So my analysis in #9 is wrong and I would like Delta² to unlike the post.

 

[Delta2]

Well many thanks @cnh1995 for running this through a simulation program (which program did you use). I have to say that this analysis is a lot more to my liking and the main reason is that the current through R1 changes (increases at start and then decreases exponentially as I thought it would do) which is a lot more in agreement with my intuition that the emf source must "understand something" of what's going on the rest of the circuit, as well as the fact that the energy to charge further C1 has to come somehow from the source (though there would be objections to this since the energy could come from C2 as well).

I unliked post 9 but I liked (really a lot) your last post so thanks again for helping me understand this.

 

[cnh1995]

which program did you use

It's an android app called EveryCircuit.

I too was skeptical about #9 because ultimately, the energy is coming from the voltage source and any change in the circuit connection must be realized by the energy source. In fact, this was my first conclusion but that was giving a decreased voltage across the C3-C4 branch, so I dropped it. The whole time I was assuming that the voltage across C3-C4 branch is not changing and is fixed at 50V. This was a major mistake and it made me think the nonsense in #9.

Anyways, the solution is clear now.